Question

question 16
let ( a ) be an invertible ( n \times n ) matrix. which of the following statements is always true?
(a) the system ( avec{x} = vec{0} ) has infinitely many solutions.
(b) ( a ) can have two identical rows.
(c) one row of ( a ) can be a multiple of another row of ( a ).
(d) the system ( avec{x} = vec{b} ) has a unique solution for all ( vec{b} in mathbb{r}^n ).
(e) the rank of ( a ) is less than ( n ).

Explanation:

Step1: Recall properties of invertible matrices

An \( n\times n \) matrix \( A \) is invertible if and only if it is row - equivalent to the \( n\times n \) identity matrix \( I_n \), and if and only if \( \det(A)
eq0 \). Also, the rank of an invertible \( n\times n \) matrix is \( n \) (full - rank), and for a linear system \( A\vec{x}=\vec{b} \), we can use the formula \( \vec{x}=A^{-1}\vec{b} \) when \( A \) is invertible.

Step2: Analyze option (A)

For the homogeneous system \( A\vec{x}=\vec{0} \), when \( A \) is invertible, we can multiply both sides of the equation by \( A^{-1} \): \( A^{-1}(A\vec{x})=A^{-1}\vec{0} \). By the associative property of matrix multiplication, \( (A^{-1}A)\vec{x}=\vec{0} \), and since \( A^{-1}A = I_n \), we have \( I_n\vec{x}=\vec{0} \), which implies \( \vec{x}=\vec{0} \). So the system \( A\vec{x}=\vec{0} \) has only the trivial solution (a unique solution), not infinitely many solutions. So option (A) is false.

Step3: Analyze option (B)

If a matrix \( A \) has two identical rows, then the determinant of \( A \) is zero (using the property that if two rows of a matrix are identical, the determinant is zero). But an invertible matrix has a non - zero determinant. So \( A \) (an invertible matrix) cannot have two identical rows. Option (B) is false.

Step4: Analyze option (C)

If one row of \( A \) is a multiple of another row, then the determinant of \( A \) is zero (using the property that if one row is a scalar multiple of another row, the determinant is zero). Since \( A \) is invertible, \( \det(A)
eq0 \), so one row of \( A \) cannot be a multiple of another row. Option (C) is false.

Step5: Analyze option (D)

For the system \( A\vec{x}=\vec{b} \), when \( A \) is invertible, we can solve for \( \vec{x} \) by multiplying both sides of the equation by \( A^{-1} \): \( \vec{x}=A^{-1}\vec{b} \). Since matrix multiplication of \( A^{-1} \) and \( \vec{b} \) gives a unique vector \( \vec{x} \) for any \( \vec{b}\in\mathbb{R}^n \), the system \( A\vec{x}=\vec{b} \) has a unique solution for all \( \vec{b}\in\mathbb{R}^n \).

Step6: Analyze option (E)

The rank of an \( n\times n \) invertible matrix \( A \) is equal to \( n \) (because the row - echelon form of \( A \) has \( n \) non - zero rows). So the rank of \( A \) is not less than \( n \), it is equal to \( n \). Option (E) is false.

Answer:

D. The system \( A\vec{x}=\vec{b} \) has a unique solution for all \( \vec{b}\in\mathbb{R}^n \)